I agree with your conclusion, but I don't follow your reasoning, and in particular how you get E = 2p/m.

We know that E (kinetic energy) = 1/2 m*v^2, and that p = mv.
If I solve the 2nd equation for v, I get v = p/m.
Substituting p/m for v in the KE equation, I get
E = 1/2 p^2/m, not 2p/m as you show.
Mark

I know we've done this one, but I wanted to show another logical way of doing this.
If we assume that we have two equal objects with the same mass and velocity and therefore momentum and kinetic energy to start.

K.E. = m*v*v
Momentum = m*v

Now let's increase the mass of one while adjusting the velocity so that the kinetic energies stay the same. So we will multiply one by the number k > 1 to get the new bigger mass.
New K.E. = (m*k)*(v*v)/k.
We have to divide by a total of k to make it stay the same. But actually each velocity will decrease by a factor of SQRT(k).

Now the new momentum will be
(m*k)*(v/SQRT(k) = m*v*SQRT(k).
Since k > 1, then SQRT(k) > 1. (Remember we are talking about magnitudes here.)

Therefore, the new momentum is larger than the old momentum. So more mass with the same kinetic energy will increase momentum.

Your algebra is wrong.. Your k multiplication would cancel out, doing nothing, there would be no square root..
Also, don't forget the 1/2 coefficient on the KE..
The simplest way to envision it is that the lighter particle's velocity advantage 'squared' merely matches the energy of the heavier particle.. When that velocity advantage is forced to compete in an equation where it doesn't get to be squared in order to compensate for its mass deficiency, it comes up short..

Your algebra is wrong.. Your k multiplication would cancel out, doing nothing, there would be no square root..
Also, don't forget the 1/2 coefficient on the KE..
The simplest way to envision it is that the lighter particle's velocity advantage 'squared' merely matches the energy of the heavier particle.. When that velocity advantage is forced to compete in an equation where it doesn't get to be squared in order to compensate for its mass deficiency, it comes up short..

Click to expand...

I will admit that it wasn't as clear as I would have like, but the algebra is correct. The k's are supposed to cancel out. Otherwise, you wouldn't maintain a constant K.E. What I'm showing is that if you increase the mass by a factor of k, in order to maintain the same kinetic energy, the velocity must be divided by SQRT(k).

No, you started with
K.E. = (m)*(v*v)
Then you multiplied m by k, and divided v*v by k.. But this is one term since everything is multiplied..
you essentially did K.E. = m*v*v*(k/k).. You multiplied (2x) the kinetic energy by 1.. I think what you were aiming at doing is multiplying both sides by K.E. in an attempt to make the actual K.E. eqv to one unit, but that isn't apropriate either..

I'm trying to make K.E. stay the same despite the fact that I am increasing mass by a factor of k. To do this, the K.E. equation must have 1/k somewhere else. Since there are only two variables, m and v, the 1/k defines the change needed in velocity. Actually it defines the change needed in v^2 so it must be a 1/SQRT(k) decrease for v.

Stated another way. We start with m0 and v0, with K.E. = m0*v0^2.
Now I define m1 = k*m0 and v1 = v0*n, where k > 1. So I am increasing mass.
If I want K.E. to stay the same in both situations, then
K.E. = m0*v0^2 = m1*v1^2 = k*m0*v0^2 * n^2
This reduces to n = 1/SQRT(k).

However, the equations are obscuring the basic idea here, instead of the original intent, which was to illustrate it better.

What I am trying to explain is how to solve problems like this using the concept of growth rates. We start with a certain situation and in a very abstract simple way, we can determine how things will relate to one another without ever really showing numbers.

In this case. If we start with a known K.E. and we increase mass by a factor of k, the velocity will decrease by a factor of SQRT(k) if we are to maintain the same K.E. Now the question is what happens to momentum. Since momentum = m*v, and m is changing by a factor of k and v is changing by 1/SQRT(k), then momentum is changing by a factor of SQRT(k). Since k > 1, SQRT(k) > 1. So as mass increases in a system maintaining the same K.E., the momentum will also increase.

Using this idea, you can show all kinds of cool things, like why animals are limited in size or why they all jump at about same height. Or you can show that as you increase the size of a boat, how the engine power scales.